Question

Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

Answer 1

We know

`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`

`thereforetan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

`=>tan^-1  1/4+tan^-1  1/5+tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

`=>tan^-1((1/4+1/5)/(1-1/4xx1/5))+tan^-1((1/5+1/6)/(1-1/5xx1/6))+tan^-1  1/x=pi/4`

`=>tan^-1((9/20)/(19/20))+tan^-1((11/30)/(29/30))+tan^-1  1/x=pi/4`

`=>tan^-1(9/19)+tan^-1(11/29)+tan^-1  1/x=pi/4`

`=>tan^-1((9/19+11/29)/(1-11/29xx1/x))+tan^-1  1/x=pi/4`

`=>tan^-1 (235/226)+tan^-1  1/x=pi/4`

`=>tan^-1((235/226+1/x)/(1-235/226xx1/x))=pi/4`

`=>(235x+226)/(226x-235)=tan  pi/4`

`=>(235x+226)/(226x-235)=1`

`=>235x+226=226x-235`

`=>9x=-461`

`=>x=-461/9`